I have a vector with structs
Struct S {
S( double a, double b ) : a_(a), b_(b) {}
double a_;
double b_;
S(S&&) = default; //ADDED
}
and I use emplace back to add instances of s to the vector
v.emplace_back( x, y );
Since it doesn't seem to be guaranteed that compilers will add a move constructor, I thought there is no harm in adding a default move constructor, so I added the line commented as ADDED.
However, this seemed to have disabled the operator= as I get the compile error
/opt/rh/devtoolset-3/root/usr/include/c++/4.9.2/bits/stl_algo.h:868: error: use of deleted function ‘S& S::operator=(const S&)’
*__result = _GLIBCXX_MOVE(*__first);
^
I dont understand why that is the case. Is my implementation of S(S&&) the wrong way to do it?
EDIT:
The error is thrown when using erase-remove
v.erase(std::remove_if(v.begin(), v.end(), deleteIf), v.end());
A default copy assignment operator will be automatically provided (performing a memberwise copy) unless the class declares a copy assignment operator, deletes the copy assignment operator, or declares a move operation. The reason for this is that the compiler assumes that if you declare any move operation, it is likely that your class holds data which should not be copied, and so it forces you to implement the copy assignment operator yourself, in an attempt to enforce the Rule of Five.
There's an excellent chart showing when default constructors and assignment operators are provided in this slideshow, slide 30.
Some older implementations of STL containers use the assignment operator when they resize the underlying containers. To resize, they allocate memory, default construct the initial objects, copy the objects over to the new container by using the assignment operator, and then release the previous objects.
On a related topic, if you are going to make an object Move-Constructible, you should also make it Move-Assignable.
Since the only data members in your struct are double's - primitive types that don't allocate resources - you don't need a move constructor for this struct.
Related
I'm trying to move things in and out from my thead safe deque:
template <typename T>
class ThreadSafeDeque
{
//..
T pop_front(void) noexcept
{
std::unique_lock<std::mutex> lock{_mutex};
while (_collection.empty())
{
_condNewData.wait(lock);
}
auto elem = std::move(_collection.front());
_collection.pop_front();
return elem;
}
private:
std::deque<T> _collection; // Concrete, not thread safe, storage.
//...
}
I created this class to insert into the Deque:
class DecodedFrame
{
public:
DecodedFrame(){}
DecodedFrame(const DecodedFrame &decodedFrame) = delete;
DecodedFrame &operator=(const DecodedFrame &) = delete;
std::unique_ptr<AVFrame, AVFrameDeleter> avFrame;
Now I'm trying to do
std::shared_ptr<ThreadSafeDeque<DecodedFrame>> decodedFramesFifo;
//add some `DecodedFrame`s to decodedFramesFifo
DecodedFrame decodedFrame = std::move(decodedFramesFifo->pop_front());
But the compiler complains that I deleted the copy assignment constructor, even though I'm trying to use the move assingment constructor. My guess is that it happens because pop_front returns T, not T&. However, returning references makes no sense becaue the object is supposed to leave the deque forever and therefore the reference to it will die.
How can I move things here?
ps: how is it possible for the compiler to copy things when the DecodedFrame holds an unique_ptr? It can't be copied!
The problem is you declared your copy c'tor and assignment operator. Doesn't matter that the declaration deletes them, it's still a user supplied declaration. That suppresses the implicit declaration of the move operations. Your options are to
Default the move operations explicitly.
Remove the copy operation declarations, they will still be deleted implicitly on account of the noncopyable member.
The copy-ctor/assign operations are deleted (these are also declarations) but that does not implicitly declare/define move-ctor/assign operations.
See p30 of https://fr.slideshare.net/ripplelabs/howard-hinnant-accu2014
You have to declare (default) them.
DecodedFrame(DecodedFrame &&) = default;
DecodedFrame &operator=(DecodedFrame &&) = default;
In order to avoid such frustrating behaviour, you should consider the rule of five.
(https://en.cppreference.com/w/cpp/language/rule_of_three#Rule_of_five)
You dont get the move constructor and move assignment operator because your copy constructor and copy assignment operator are user delcared/defined (you deleted them). you can force the default move constructor and move assignment via "=default" (like you did with delete).
But because the class has a unique pointer as member which itself is only move constructible and move assignable, you will get the deletion of the copy constructor and copy assignment for free. just remove you delete statements and you will be fine, since you then again get the move operations.
The title pretty much sums up my question. In more detail: I know that when I declare a move constructor and a move assignment operator in C++11 I have to "make the other objects variables zero". But how does that work, when my variable is not an array or a simple int or double value, but its a more "complex" type?
In this example I have a Shoplist class with a vector member variable. Do I have to invoke the destructor of the vector class in the move assignment operator and constructor? Or what?
class Shoplist {
public:
Shoplist() :slist(0) {};
Shoplist(const Shoplist& other) :slist(other.slist) {};
Shoplist(Shoplist&& other) :slist(0) {
slist = other.slist;
other.slist.~vector();
}
Shoplist& operator=(const Shoplist& other);
Shoplist& operator=(Shoplist&& other);
~Shoplist() {};
private:
vector<Item> slist;
};
Shoplist& Shoplist::operator=(const Shoplist& other)
{
slist = other.slist;
return *this;
}
Shoplist& Shoplist::operator=(Shoplist&& other)
{
slist = other.slist;
other.slist.~vector();
return *this;
}
Whatever a std::vector needs to do in order to move correctly, will be handled by its own move constructor.
So, assuming you want to move the member, just use that directly:
Shoplist(Shoplist&& other)
: slist(std::move(other.slist))
{}
and
Shoplist& Shoplist::operator=(Shoplist&& other)
{
slist = std::move(other.slist);
return *this;
}
In this case, you could as AndyG points out, just use = default to have the compiler generate exactly the same move ctor and move assignment operator for you.
Note that explicitly destroying the original as you did is definitely absolutely wrong. The other member will be destroyed again when other goes out of scope.
Edit: I did say assuming you want to move the member, because in some cases you might not.
Generally you want to move data members like this if they're logically part of the class, and much cheaper to move than copy. While std::vector is definitely cheaper to move than to copy, if it holds some transient cache or temporary value that isn't logically part of the object's identity or value, you might reasonably choose to discard it.
Implementing copy/move/destructor operations doesn't make sense unless your class is managing a resource. By managing a resource I mean be directly responsible for it's lifetime: explicit creation and destruction. The rule of 0 and The rule of 3/5 stem from this simple ideea.
You might say that your class is managing the slist, but that would be wrong in this context: the std::vector class is directly (and correctly) managing the resources associated with it. If you let our class have implicit cpy/mv ctos/assignment and dtors, they will correctly invoke the corresponding std::vector operations. So you absolutely don't need to explicitly define them. In your case the rule of 0 applies.
I know that when I declare a move constructor and a move assignment
operator in C++11 I have to "make the other objects variables zero"
Well no, not really. The ideea is that when you move from an object (read: move it's resource from an object) then you have to make sure that your object it's left aware that the resource it had is no more under it's ownership (so that, for instance, it doesn't try to release it in it's destructor). In the case of std::vector, it's move ctor would set the pointer it has to the internal buffer to nullptr.
I know that when I declare a move constructor and a move assignment operator in C++11 I have to "make the other objects variables zero"
This is not quite correct. What you must do, is maintain validity of the moved from object. This means that you must satisfy the class invariant.
If you have specified a special invariant for a particular class, that requires you to set member variables to zero, then perhaps such class might have to do so. But this is not a requirement for move in general.
Do I have to invoke the destructor of the vector class in the move assignment operator and constructor?
Definitely not. The destructors of the members will be called when the moved from object is destroyed.
What you would typically do, is move construct/assign each member in the move constructor/assignment operator of the containing object. This is what the implicitly generated special member functions do. Of course, this might not satisfy the class invariant for all classes, and if it doesn't, then you may need to write your own versions of them.
The compiler will implicitly generate the special member functions for you, if you don't try to declare them yourself. Here is a minimal, but correct version of your class:
class Shoplist {
vector<Item> slist;
};
This class is default constructible, movable and copyable.
The move constructor should move member-wise:
Shoplist(Shoplist&& other)
: slist(std::move(other.slist))
{}
Note, that the compiler generates move constructors for you (when possible) by member-wise move, as you would do by hand above.
Move constructors are allowed (but not required) "steal" the contents of the moved-from object. This does not mean that they must "make the other objects variables zero". Moving a primitive type, for instance, is equivalent to copying it. What it does mean is that a move constructor can transfer ownership of data in the heap or free store. In this case, the moved-from object must be modified so that when it is destroyed (which should not happen in the move-constructor), the data it previously owned (before it was transferred) will not be freed.
Vector provides its own move constructor. So all you need to do in order to write a correct move constructor for an object containing a vector is to ensure the correct vector constructor is invoked. This is done by explicitly passing an r-value reference to the sub-object constructor, using std::move:
Shoplist(Shoplist&& other) :slist(std::move(other.slist)) {
//... Constructor body
... But in fact you probably don't need to do this in general. Your copy and move constructors will be correctly auto-generated if you don't declare them and don't declare a destructor. (Following this practice is called the "rule of 0".)
Alternatively, you can force the compiler to auto-generate the move constructor:
Shoplist(Shoplist&& other) = default;
My class A explicitly implements both its copy constructor and its copy assignment.
Which copy mechanism is used when copy assigning a vector of such elements?
Is this:
vector<A> a1, a2(5);
a1 = a2;
going to use A's copy constructor for all new elements of a1, with the elements of a2 as input?
Or is it going to make room in a1 for the elements, then use A's operator= with the elements of a2 as input?
What if a1 isn't empty before the assignment?
Is it even specified?
My class's copy constructor and operator= don't exactly do the same thing (is it bad practice? Mainly testing stuff so far). It looks like the copy constructor is called, but I wonder if it is guaranteed to be that way or if it just happens to be so in this very case.
In this context it will call the copy constructor 5 times. Since a1 is empty, there aren't any elements to assign to. So they need to be copy-constructed.
Generally, it will call whatever mixture of copy/move construction/assignment or deletion is appropriate. All depending on the sizes of the vectors in question, the particular vector operation you are performing, and the value categories of the operands.
It looks like the copy constructor is called, but I wonder if it is guaranteed to be that way or if it just happens to be so in this very case.
It could be guaranteed for the exception-safe swap-based implementation of assignment:
struct SomeClass
{
SomeClass(const SomeClass& other) { ... }
SomeClass(SomeClass&& other) { ... }
// Copy/move construction is performed while initializing the parameter
void operator=(SomeClass other)
{
this->swap(other);
}
void swap(SomeClass& other) { ... }
};
The disadvantage of such an implementation of assignment is that - because of its generality - it is not the most optimal one (for example, it does unnecessary job in case of self assignment).
In general, if exception safety concerns can be rules out, copy assigning to an object can be done faster than destroying it and constructing an in-place copy of the source object. Therefore, you should expect that performance-seeking implementations would perform assignment of objects through assignment rather than copy-construction of their sub-objects wherever possible.
I recently revisited the copy constructor, assignment operator, copy swap idom seen here:
What is the copy-and-swap idiom?
and many other places -
The Above link is an excellent post - but I still had a few more questions -
These questions are answered in a bunch of places, on stackoverflow and many other sites, but I have not seen a lot of consistency -
1 - Should you have try-catch around the areas where we allocate the new memory for a deep copy in the copy constructor ? (Ive seen it both ways)
2 - With regards to inheritance for both the copy constructor and assignment operator, when should the base class functions be called, and when should these functions be virtual?
3 - Is std::copy the best way for duplicating memory in the copy constructor? I have seen it with memcpy, and seen others say memcpy the worst thing on earth.
Consider the example Below (Thanks for all the feedback), it prompted some additional questions:
4 - Should we be checking for self assignment? If so where
5 - Off topic question, but I have seen swapped used as :
std::copy(Other.Data,Other.Data + size,Data);
should it be:
std::copy(Other.Data,Other.Data + (size-1),Data);
if swap goes from 'First to Last' and the 0th element is Other.Data?
6 - Why doesn't the commented out constructor work (I had to change size to mysize) - is assume this means regardless of the order I write them, the constructor will always call the allocation element first?
7 - Any other comments on my implementation? I know the code is useless but i'm just trying to illustrate a point.
class TBar
{
public:
//Swap Function
void swap(TBar &One, TBar &Two)
{
std::swap(One.b,Two.b);
std::swap(One.a,Two.a);
}
int a;
int *b;
TBar& operator=(TBar Other)
{
swap(Other,*this);
return (*this);
}
TBar() : a(0), b(new int) {} //We Always Allocate the int
TBar(TBar const &Other) : a(Other.a), b(new int)
{
std::copy(Other.b,Other.b,b);
*b = 22; //Just to have something
}
virtual ~TBar() { delete b;}
};
class TSuperFoo : public TBar
{
public:
int* Data;
int size;
//Swap Function for copy swap
void swap (TSuperFoo &One, TSuperFoo &Two)
{
std::swap(static_cast<TBar&>(One),static_cast<TBar&>(Two));
std::swap(One.Data,Two.Data);
std::swap(One.size,Two.size);
}
//Default Constructor
TSuperFoo(int mysize = 5) : TBar(), size(mysize), Data(new int[mysize]) {}
//TSuperFoo(int mysize = 5) : TBar(), size(mysize), Data(new int[size]) {} *1
//Copy Constructor
TSuperFoo(TSuperFoo const &Other) : TBar(Other), size(Other.size), Data(new int[Other.size]) // I need [Other.size]! not sizw
{
std::copy(Other.Data,Other.Data + size,Data); // Should this be (size-1) if std::copy is First -> Last? *2
}
//Assignment Operator
TSuperFoo& operator=(TSuperFoo Other)
{
swap(Other,(*this));
return (*this);
}
~TSuperFoo() { delete[] Data;}
};
If you allocate memory then you need to ensure that it is freed in the case of an exception being thrown. You can do this with an explicit try/catch, or you can use a smart pointer such as std::unique_ptr to hold the memory, which will then be automatically deleted when the smart pointer is destroyed by stack unwinding.
You very rarely need a virtual assignment operator. Call the base class copy constructor in the member initialization list, and base-class assignment operator first in the derived assignment operator if you are doing a memberwise assignment --- if you are doing copy/swap then you don't need to call the base class assignment in your derived assignment operator, provided that copy and swap are implemented correctly.
std::copy works with objects, and will correctly call copy constructors. If you have plain POD objects then memcpy will work just as well. I'd go for std::copy in most cases though --- it should be optimized to memcpy under the hood anyway for PODs, and it avoids the potential for errors should you add a copy constructor later.
[Updates for updated question]
With copy/swap as written there is no need to check for self-assignment, and indeed no way of doing so --- by the time you enter the assignment operator other is a copy, and you have no way of knowing what the source object was. This just means that self-assignment will still do a copy/swap.
std::copy takes a pair of iterators (first, first+size) as input. This allows for empty ranges, and is the same as every range-based algorithm in the standard library.
The commented out constructor doesn't work because the members are initialized in the order they are declared, regardless of the order in the member initializer list. Consequently, Data is always initialized first. If the initialization depends on size then it will get a duff value since size hasn't been initialized yet. If you swap the declarations of size and data then this constructor will work fine. Good compilers will warn about the order of member initialization not matching the order of declarations.
1 - Should you have try-catch around the areas where we allocate the new memory for a deep copy in the copy constructor ?
In general, you should only catch an exception if you can handle it. If you have a way of dealing with an out-of-memory condition locally, then catch it; otherwise, let it go.
You should certainly not return normally from a constructor if construction has failed - that would leave the caller with an invalid object, and no way to know that it's invalid.
2 - With regards to inheritance for both the copy constructor and assignment operator, when should the base class functions be called, and when should these functions be virtual?
A constructor can't be virtual, since virtual functions can only be dispatched by an object, and there is no object before you create it. Usually, you wouldn't make assignment operators virtual either; copyable and assignable classes are usually treated as non-polymorphic "value" types.
Usually, you'd call the base class copy constructor from the initialiser list:
Derived(Derived const & other) : Base(other), <derived members> {}
and if you're using the copy-and-swap idiom, then your assignment operator wouldn't need to worry about the base class; that would be handled by the swap:
void swap(Derived & a, Derived & b) {
using namespace std;
swap(static_cast<Base&>(a), static_cast<Base&>(b));
// and swap the derived class members too
}
Derived & Derived::operator=(Derived other) {
swap(*this, other);
return *this;
}
3 - Is std::copy the best way for duplicating memory in the copy constructor? I have seen it with memcopy, and seen others say memcopy the worst thing on earth.
It's rather unusual to be dealing with raw memory; usually your class contains objects, and often objects can't be correctly copied by simply copying their memory. You copy objects using their copy constructors or assignment operators, and std::copy will use the assignment operator to copy an array of objects (or, more generally, a sequence of objects).
If you really want, you could use memcpy to copy POD (plain old data) objects and arrays; but std::copy is less error-prone (since you don't need to provide the object size), less fragile (since it won't break if you change the objects to be non-POD) and potentially faster (since the object size and alignment are known at compile time).
If the constructor for what you're deep copying may throw something
you can handle, go ahead and catch it. I'd just let memory
allocation exceptions propagate, though.
Copy constructors (or any constructors) can't be virtual. Include a
base class initializer for these. Copy assignment operators should
delegate to the base class even if they're virtual.
memcpy() is too low-level for copying class types in C++ and can lead to undefined behavior. I think std::copy is usually a better choice.
try-catch can be used when you have to undo something. Otherwise, just let the bad_alloc propagate to the caller.
Calling the base class' copy constructor or assignment operator is the standard way of letting is handle its copying. I have never seen a use case for a virtual assignment operator, so I guess they are rare.
std::copy has the advantage that it copies class objects correctly. memcpy is rather limited on what types it can handle.
If the operator= is properly defined, is it OK to use the following as copy constructor?
MyClass::MyClass(MyClass const &_copy)
{
*this = _copy;
}
If all members of MyClass have a default constructor, yes.
Note that usually it is the other way around:
class MyClass
{
public:
MyClass(MyClass const&); // Implemented
void swap(MyClass&) throw(); // Implemented
MyClass& operator=(MyClass rhs) { rhs.swap(*this); return *this; }
};
We pass by value in operator= so that the copy constructor gets called. Note that everything is exception safe, since swap is guaranteed not to throw (you have to ensure this in your implementation).
EDIT, as requested, about the call-by-value stuff: The operator= could be written as
MyClass& MyClass::operator=(MyClass const& rhs)
{
MyClass tmp(rhs);
tmp.swap(*this);
return *this;
}
C++ students are usually told to pass class instances by reference because the copy constructor gets called if they are passed by value. In our case, we have to copy rhs anyway, so passing by value is fine.
Thus, the operator= (first version, call by value) reads:
Make a copy of rhs (via the copy constructor, automatically called)
Swap its contents with *this
Return *this and let rhs (which contains the old value) be destroyed at method exit.
Now, we have an extra bonus with this call-by-value. If the object being passed to operator= (or any function which gets its arguments by value) is a temporary object, the compiler can (and usually does) make no copy at all. This is called copy elision.
Therefore, if rhs is temporary, no copy is made. We are left with:
Swap this and rhs contents
Destroy rhs
So passing by value is in this case more efficient than passing by reference.
It is more advisable to implement operator= in terms of an exception safe copy constructor. See Example 4. in this from Herb Sutter for an explanation of the technique and why it's a good idea.
http://www.gotw.ca/gotw/059.htm
This implementation implies that the default constructors for all the data members (and base classes) are available and accessible from MyClass, because they will be called first, before making the assignment. Even in this case, having this extra call for the constructors might be expensive (depending on the content of the class).
I would still stick to separate implementation of the copy constructor through initialization list, even if it means writing more code.
Another thing: This implementation might have side effects (e.g. if you have dynamically allocated members).
While the end result is the same, the members are first default initialized, only copied after that.
With 'expensive' members, you better copy-construct with an initializer list.
struct C {
ExpensiveType member;
C( const C& other ): member(other.member) {}
};
};
I would say this is not okay if MyClass allocates memory or is mutable.
yes.
personally, if your class doesn't have pointers though I'd not overload the equal operator or write the copy constructor and let the compiler do it for you; it will implement a shallow copy and you'll know for sure that all member data is copied, whereas if you overload the = op; and then add a data member and then forget to update the overload you'll have a problem.
#Alexandre - I am not sure about passing by value in assignment operator. What is the advantage you will get by calling copy constructor there? Is this going to fasten the assignment operator?
P.S. I don't know how to write comments. Or may be I am not allowed to write comments.
It is technically OK, if you have a working assignment operator (copy operator).
However, you should prefer copy-and-swap because:
Exception safety is easier with copy-swap
Most logical separation of concerns:
The copy-ctor is about allocating the resources it needs (to copy the other stuff).
The swap function is (mostly) only about exchanging internal "handles" and doesn't need to do resource (de)allocation
The destructor is about resource deallocation
Copy-and-swap naturally combines these three function in the assignment/copy operator